Biggest danger I can see with a break to SHA or any signing algorithm is
just the danger of masquerading and pretending something is coming from
someone it's not. i.e. the algorithim becomes worthless as a way of
establishing identity of the signer.
Of course you can always tell which one is the evil twin because he/she
will be the one with the goatee.
Unless of course the sender is s robot named Bender in which case the good
twin is the one with the goatee.
For something like bitcoin a break to the signing algorithm would be
devastating since anyone with the key has the authority to spend the
coins. Although it uses ecdsa not SHA.
As far as what does being able to solve it in 1/2 an operation really mean,
well of course it's a quantum computer so it means that it would be both
solved and unsolved at the same time. :)
On Fri, Jun 21, 2013 at 5:14 PM, Jason Klebs <jasonk at riseup.net> wrote:
> Not being quite too familiar with it...what exactly does it mean to
> break SHA256?
>> I understand that SHA is a message digest, and that
> SHA256(<arbitrary-length input>) = fixed-length digest.
>> I also understand that, for example, digital signatures for public key
> encryption (like SSL/TLS) utilize such message digests. Does "breaking"
> SHA mean that the breaker can generate and sign keys without the
> corresponding RSA private key? Does it mean that the breaker has found
> any and all inputs where SHA(input) = digest? Something else?
>> -Jason
>> On 06/21/2013 05:02 PM, Todd Millecam wrote:
> > The architecture might be higher, but the limiting factor on that is how
> > many quantum entangled pairs they can emit--and the max is still 13
> > whatever their specs docs say, so no, it can't break sha in one
> operation.
> >
> >
>>>> /*
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